Question

A box P and a coil Q are connected in series with an ac source of variable frequency. The emf of the source is constant at $10\text{V}$. Box P contains a capacitance of $1\mu F$ in series with a resistance of $32\Omega$. Coil Q has a self inductance of 4.9 mH and a resistance of $68\Omega$ in series. The frequency is adjusted so that maximum current flows in P and Q.

The voltage across Q is

• A. $20\text{V}$
• B. $7.7\text{V}$
• C. $10\text{V}$
• D. $\frac{\sqrt{9524}}{10}\text{V}$

Answer 1

The correct option is D $\frac{\sqrt{9524}}{10}\text{V}$

Voltage across Q can be given as
$V_Q=I_{rms}\times Z_Q$ .....(i)
For impedence in coil Q
$Z=\sqrt{R_2^2+(X_L{)}^2}$ ....(ii)
Now if the maximum current flows through P and Q, then circuit is at resonance. Also at resonance $\omega =\frac{1}{\sqrt{LC}}$
$\omega =\frac{1}{\sqrt{4.9\times {10}^{-3}\times {10}^{-6}}}\omega =\frac{{10}^5}{7}$
So we can find $X_L$ as
$X_L=\omega L=\frac{{10}^5}{7}\times 4.9\times {10}^{-3}=70\Omega$
Putting values in eq(i)
$Z=\sqrt{{68}^2+{70}^2}=\sqrt{9524}\Omega$ ......(iii)
To calulate $I_{rms}$ we need to understand that the circuit is at resonance and it behaves as purely resistive ciruit.
So
$I_{rms}=\frac{V_{rms}}{R_1+R_2}=\frac{10}{68+32}=0.1A$ .....(iv)
From (iii) and (iv)
$V_P=0.1\times \sqrt{9524}=\frac{\sqrt{9524}}{10}\text{V}$